Question: The differentiable functions $x$ and $y$ are related by the following equation: $y=\tan(x)$ Also, $\dfrac{dy}{dt}=-6$. Find $\dfrac{dx}{dt}$ when $x=-\dfrac{\pi}{3}$.
Explanation: Let's start by differentiating the equation $y=\tan(x)$ with respect to $t$. $\begin{aligned} y&=\tan(x) \\\\ \dfrac{dy}{dt}&=\sec^2(x)\cdot\dfrac{dx}{dt} \end{aligned}$ We are given that $\dfrac{dy}{dt}=-6$, and we want to find $\dfrac{dx}{dt}$ when $x=-\dfrac{\pi}{3}$. Let's plug ${x=-\dfrac{\pi}{3}}$ and ${\dfrac{dy}{dt}=-6}$ into the equation we obtained: $\begin{aligned} {\dfrac{dy}{dt}}&=\sec^2({x})\cdot\dfrac{dx}{dt} \\\\ {-6}&=\sec^2\!\left({-\dfrac{\pi}{3}}\right)\cdot\dfrac{dx}{dt} \\\\ -6&=4\cdot\dfrac{dx}{dt} \\\\ -1.5&=\dfrac{dx}{dt} \end{aligned}$ In conclusion, when $x=-\dfrac{\pi}{3}$, the value of $\dfrac{dx}{dt}$ is $-1.5$.